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xz_007 [3.2K]
2 years ago
15

If vertex A is at (-1, 2) and vertex B is at (1, 5), then vertex A' is at and vertex B' is at .

Mathematics
1 answer:
Serggg [28]2 years ago
8 0

Answer:

Vertical translation is along the y axis while horizontal translation is along the x-axis.

Vertex A' is at (-1 - 3, 2 + 6) = (-4, 8)

Vertex B' is at (1 - 3, 5 + 6) = (-2, 11)

Brainliest Pls

Step-by-step explanation:

You might be interested in
Which calculation will ALWAYS give a result greater than 1? A.5 × a number less than 1 B.4/9+ a fraction less than 12 C. 1 3/4 -
Bas_tet [7]

Answer:

  C.  1 3/4 - (a fraction less than 3/4)

Step-by-step explanation:

Your number sense should be able to help you with this one.

A. 5 × 1/10 = 1/2, not a number greater than 1

B. 4/9 + 1/3 = 7/9, not a number greater than 1

C. 1 3/4 -1/2 = 1 1/4, a number greater than 1 (see below for more explanation)

D. 7/8 × 1/2 = 7/16, not a number greater than 1.

__

<em>More explanation</em>

Let x represent a number less than 3/4. Then we want to make sure that ...

  y = 1 3/4 - x

will be greater than 1.

Solving for x, we get ...

  x = 1 3/4 - y

Applying the requirement that x < 3/4, we have ...

  x < 3/4

  (1 3/4 -y) < 3/4

  1 3/4 < y + 3/4 . . . . . . add y

  1 < y . . . . . . . . . . . . . . .subtract 3/4

We see that the condition on x makes sure that y is always greater than 1.

3 0
3 years ago
Find the roots of h(t) = (139kt)^2 − 69t + 80
Sonbull [250]

Answer:

The positive value of k will result in exactly one real root is approximately 0.028.

Step-by-step explanation:

Let h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80, roots are those values of t so that h(t) = 0. That is:

19321\cdot k^{2}\cdot t^{2}-69\cdot t + 80=0 (1)

Roots are determined analytically by the Quadratic Formula:

t = \frac{69\pm \sqrt{4761-6182720\cdot k^{2} }}{38642}

t = \frac{69}{38642} \pm \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321}  }

The smaller root is t = \frac{69}{38642} - \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321}  }, and the larger root is t = \frac{69}{38642} + \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321}  }.

h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80 has one real root when \frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} = 0. Then, we solve the discriminant for k:

\frac{80\cdot k^{2}}{19321} = \frac{4761}{1493204164}

k \approx \pm 0.028

The positive value of k will result in exactly one real root is approximately 0.028.

7 0
2 years ago
There is 1/4 ounce of yeast in every 2 1/4 teaspoon of yeast. A recipe calls for 2 teaspoons of yeast. How many ounces of yeast
MrRissso [65]
1/4 ounce of yeast ... 2 1/4 teaspoons of yeast
x ounce of yeast = ? ... 2 teaspoons of yeast

If you would like to know how many ounces of yeast need to be in the recipe, you can calculate this using the following steps:

1/4 * 2 = x * 2 1/4
1/2 = x * 9/4     /*4/9
x = 1/2 * 4/9
x = 2/9

Result: 2/9 ounce of yeast needs to be in this recipe.
7 0
3 years ago
5^(2x-1)+5^(x+1)=250<br> how do you solve?<br> thank you
zepelin [54]

Answer:

<em>x = 2</em>

Step-by-step explanation:

<u>Exponential Equations</u>

Solve:

5^{2x-1}+5^{x+1}=250

Separate each exponential:

5^{2x}5^{-1}+5^{x}5^{1}=250

Operating:

\displaystyle \frac{5^{2x}}{5}+5^{x}5=250

Multiplying by 5:

5^{2x}+25\cdot5^x=1250

Rearranging:

5^{2x}+25\cdot5^x-1250=0

Recall that:

5^{2x}=(5^{x})^2

(5^{x})^2+25\cdot5^x-1250=0

Calling

y=5^{x}:

y^2+25y-1250=0

Factoring:

(y-25)(y+50)=0

There are two possible solutions:

y=25

y=-50

Since

y=5^{x}

y cannot be negative, thus:

5^{x}=25=5^2

The solution is:

x = 2

8 0
3 years ago
Solve for x. 3x + 5 = 23​
Lunna [17]

[|] Answer [|]

\boxed{X \ = \ 6}

[|] Explanation [|]

3x + 5 = 23

_________

_________

Subtract 5 From Both Sides:

3x + 5 - 5 = 23 - 5

Simplify:

3x = 18

Divide Both Sides By 3:

\frac{3x}{3} \ = \ \frac{18}{3}

Simplify:

X = 6

_________

_________

- Check Your Work -

Substitute 6 For X:

3 * 6 + 5 = 23

Parenthesis

Exponents

Multiply

Divide

Add

Subtract

3 * 6 = 18

18 + 5 = 23

\boxed{[|] \ Eclipsed \ [|]}

8 0
3 years ago
Read 2 more answers
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